STA3710 ASSIGNMENT 4 2025 (894289)

STA3710
ASSIGNMENT 4 2025

UNIQUE NO. 894289
DUE DATE: 9 SEPTEMBER 2025

, Mathematical Techniques in Statistics


Question 1.1

Let 𝐴 be partitioned as

𝐴 𝐴12
𝐴 = ൬ 11 ൰
,
𝐴21 𝐴22

with 𝐴11 of size 𝑚1 × 𝑚1 and rank(𝐴) = rank(𝐴11 ) = 𝑚1 . Show that

−1
𝐴22 = 𝐴21 𝐴11 𝐴12 .

Solution. rank(𝐴11 ) = 𝑚1 means 𝐴11 is full column-rank (in fact square and invertible),
−1
so 𝐴11 exists.

Consider the column space of 𝐴. Write 𝐴 as block-column form:

𝐴 𝐴
𝐴 = ൬ 11 ൰concatenated with ൬ 12 ൰
.
𝐴21 𝐴22

𝐴
Since rank(𝐴) = rank(𝐴11 ) = 𝑚1 , every column of the second block-column ൬ 12 ൰must
𝐴22
𝐴
be a linear combination of the columns of the first block-column ൬ 11 ൰
. Therefore there
𝐴21
exists a matrix 𝑋 (of appropriate size) such that

𝐴 𝐴
൬ 12 ൰= ൬ 11 ൰𝑋.
𝐴22 𝐴21

Looking at the top block gives 𝐴12 = 𝐴11 𝑋. Since 𝐴11 is invertible,

−1
𝑋 = 𝐴11 𝐴12 .

Then the bottom block gives

−1
𝐴22 = 𝐴21 𝑋 = 𝐴21 𝐴11 𝐴12 ,

which is what we wanted to show. ▫